AP Calculus AB and BC

U-Substitution vs Integration by Parts

Q: How do I choose u for integration by parts?

Pick u so that differentiating it simplifies the problem, and dv so that you can actually integrate it. A polynomial factor is usually a good u because repeated differentiation eventually kills it.

Use substitution when the integrand contains a function and something close to its derivative, since substitution reverses the chain rule. Use integration by parts when the integrand is a product of two unlike types, such as a polynomial times an exponential, since parts reverses the product rule.

U-substitution

Use when: You can spot an inner function whose derivative is also present, up to a constant factor.

Integration by parts

Use when: The integrand multiplies two different kinds of function and no inner derivative is available.

Side by side

	U-substitution	Integration by parts
Reverses	The chain rule	The product rule
Look for	$f(g(x)) \cdot g'(x)$	A product of unlike types
Formula	$\int f(u)\,du$ after substituting	$\int u\,dv = uv - \int v\,du$
Example	$\int 2x\cos(x^2)\,dx$	$\int x e^x\,dx$

Try substitution first, because it is faster and because failing costs nothing. If no candidate for $u$ leaves you with its own derivative sitting in the integrand, substitution stalls and parts becomes the next move.

The two examples show the difference clearly. In $\int 2x\cos(x^2)\,dx$ the factor $2x$ is exactly the derivative of $x^2$ , so substitution collapses it. In $\int x e^x\,dx$ neither factor is the derivative of the other, so parts is the way through.

Changing the bounds

For a definite integral, substitution requires you either to convert the bounds to the new variable or to convert back before evaluating. Mixing old bounds with a new variable is a routine and costly error.

Frequently asked questions

How do I choose u for integration by parts?

Pick $u$ so that differentiating it simplifies the problem, and $dv$ so that you can actually integrate it. A polynomial factor is usually a good $u$ because repeated differentiation eventually kills it.

In the CED: Unit 6: Integration and Accumulation